Minimal Polynomial

Hermitian Matrices

Definition: An n×nn \times n matrix AA is said to be Hermitian if A=AH=AA = A^H = A^*. Its conjugate transpose is equal to itself. If AA is real, then A=ATA = A^T.


Theorem: Let AA be Hermitian. Then x,Ax \langle x,Ax \rangle is real for all xCnx \in \mathbb{C}^n.

Proof: We will start with properties of inner products.

1- x,y=y,x \large \langle x,y\rangle = \overline{\langle y,x\rangle }.

Then we will substitute A=AA = A^* and in the last step we will use,

2- x,Py=Px,y \large \langle x,Py\rangle = \langle P^*x,y\rangle .

Check the properties of inner products here.

x,Ax=Ax,x=Ax,x=x,Ax     \large\langle x,Ax\rangle = \overline{\langle Ax,x\rangle } = \overline{\langle A^*x,x\rangle } = \overline{\langle x,Ax\rangle } \ \ \ \ \ \blacksquare


Theorem: Let AA be Hermitian. Then all eigenvalues of AA are real.

Proof: Let λ\lambda be an eigenvalue of AA and xx be the corresponding eigenvector. Then Ax=λxAx = \lambda x. Then we will use the previous theorem.

λx,x=λx,x=Ax,x=x,Ax=x,λx=λx,x=λx,x\large\lambda\langle x,x\rangle = \langle \lambda x,x\rangle = \langle Ax,x\rangle = \overline{\langle x,Ax\rangle } = \overline{\langle x,\lambda x\rangle } = \overline{\lambda\langle x,x\rangle } = \overline{\lambda}\langle x,x\rangle

Since x,x0\langle x,x\rangle \neq 0, we can divide both sides by x,x\langle x,x\rangle .

λ=λ     \large\lambda = \overline{\lambda} \ \ \ \ \ \blacksquare


Theorem: Let AA be Hermitian. Then all eigenvectors corresponding to distinct eigenvalues are orthogonal. Let AA be Hermitian and λiλj\lambda_i \neq \lambda_j be two distinct eigenvalues of AA with corresponding eigenvectors eie_i and eje_j. Then ei,ej=0\langle e_i,e_j\rangle = 0.

Proof: Let AA be Hermitian and λiλj\lambda_i \neq \lambda_j be two distinct eigenvalues of AA with corresponding eigenvectors eie_i and eje_j. Then Aei=λieiAe_i = \lambda_i e_i and Aej=λjejAe_j = \lambda_j e_j. Then we will use the previous theorem.

ei,Aej=ei,λjej=λjei,ejei,Aej=Aei,ej=λiei,ej\large \langle e_i,Ae_j\rangle = \langle e_i,\lambda_je_j \rangle = \lambda_j \langle e_i,e_j\rangle \\ \large \langle e_i,Ae_j\rangle = \langle A^*e_i,e_j\rangle =\lambda_i \langle e_i,e_j\rangle

Then we will subtract the equations.

(λjλi)ei,ej=0\large (\lambda_j - \lambda_i)\langle e_i,e_j\rangle = 0

Since λjλi \large \lambda_j \neq \lambda_i, we have ei,ej=0      \large \langle e_i,e_j\rangle = 0 \ \ \ \ \ \blacksquare


Theorem: Let AA be Hermitian. Then its minimal polynomial is

m(s)=(sλ1)(sλ2)(sλσ) \large m(s) = (s - \lambda_1 ) (s - \lambda_2 ) \cdots (s - \lambda_{\sigma} )

where λi\lambda_i are the distinct eigenvalues of AA.

Proof: Set Equality We will prove that m(s)m(s) has no repeated roots. For which we need to use N((AλI))=N((AλI)2)N((A-\lambda I)) = N((A-\lambda I)^2).

In order to show the equality, we will prove that N((AλI))N((AλI)2)N((A-\lambda I)) \subseteq N((A-\lambda I)^2) and N((AλI)2)N((AλI))N((A-\lambda I)^2) \subseteq N((A-\lambda I)).

1- N((AλI))N((AλI)2)N((A-\lambda I)) \subseteq N((A-\lambda I)^2)

Let xN((AλI))x \in N((A-\lambda I)). Then (AλI)x=0(A-\lambda I) x = 0. Then we will multiply both sides by (AλI)(A-\lambda I).

(AλI)2x=0(A-\lambda I)^2 x = 0

Then xN((AλI)2)x \in N((A-\lambda I)^2).

2- N((AλI)2)N((AλI))N((A-\lambda I)^2) \subseteq N((A-\lambda I))

Let xN((AλI)2)x \in N((A-\lambda I)^2). Then (AλI)2x=0(A-\lambda I)^2 x = 0.

x,(AλI)2x=0 \langle x, (A-\lambda I)^2 x \rangle = 0

x,(AλI)(AλI)x=0 \langle x, (A-\lambda I) (A-\lambda I) x \rangle = 0

(AλI)x,(AλI)x=0=(AλI)x2    (AλI)x=0 \langle (A-\lambda I) x, (A-\lambda I) x \rangle = 0 = \|(A-\lambda I)x\|^2 \implies (A-\lambda I)x = 0

Then xN((AλI))x \in N((A-\lambda I)).

Therefore Hermitian matrices AA with distinct eigenvalues have no repeated roots in their minimal polynomials. \blacksquare

d(s)=(sλ1)(sλ2)(sλσ) \large d(s) = (s - \lambda_1 ) (s - \lambda_2 ) \cdots (s - \lambda_{\sigma} )
m(s)=(sλ1)(sλ2)(sλσ) \large m(s) = (s - \lambda_1 ) (s - \lambda_2 ) \cdots (s - \lambda_{\sigma} )
Cn=N((Aλ1I))N((Aλ2I))N((AλσI)) \large \mathbb{C}^n = N((A-\lambda_1 I)) \oplus N((A-\lambda_2 I)) \oplus \cdots \oplus N((A-\lambda_{\sigma} I))

Theorem: Let AA be Hermitian matrix with characteristic polynomial d(s)=(sλ1)r1(sλ2)r2(sλσ)rσd(s) = (s - \lambda_1 )^{r_1} (s - \lambda_2 )^{r_2} \cdots (s - \lambda_{\sigma} )^{r_{\sigma}}. Then there exist a unitary matrix PP such that P1=PP^{-1} = P^* and PAP=ΛP^*AP = \Lambda where Λ\Lambda is a diagonal matrix with diagonal entries λ1,λ2,,λσ\lambda_1, \lambda_2, \cdots, \lambda_{\sigma}.

Λ=[λ10000λ20000λ30000λσ] and each ΛiCri×ri , Λi=[λi0000λi0000λi0000λi] \large \Lambda = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & 0 \\ 0 & 0 & \lambda_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_\sigma \end{bmatrix} \text{ and each }\Lambda_i \in \mathbb{C}^{r_i \times r_i} \text{ , } \Lambda_i = \begin{bmatrix} \lambda_i & 0 & 0 & \cdots & 0 \\ 0 & \lambda_i & 0 & \cdots & 0 \\ 0 & 0 & \lambda_i & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_i \end{bmatrix}

where λi\lambda_i are the distinct eigenvalues of AA.

Proof: See the lecture notes, page 50.


Theorem: Let AA be Hermitian matrix with eigenvalues λ1,λ2,,λσ\lambda_1, \lambda_2, \cdots, \lambda_{\sigma}. Let λmin=mini=1,2,,σλi\lambda_min = \min_{i=1,2,\cdots,\sigma} \lambda_i and λmax=maxi=1,2,,σλi\lambda_max = \max_{i=1,2,\cdots,\sigma} \lambda_i. Then for all xCnx \in \mathbb{C}^n,

λminx,xx,Axλmaxx,x \large \lambda_{min} \langle x,x \rangle \leq \langle x,Ax \rangle \leq \lambda_{max} \langle x,x \rangle

Proof: See the lecture notes, page 51.


Definition: A Hermitian matrix AA is said to be positive definite if x,Ax>0\langle x,Ax \rangle > 0 for all xCnx \in \mathbb{C}^n and x0x \neq 0. A Hermitian matrix AA is said to be positive semidefinite if x,Ax0\langle x,Ax \rangle \geq 0 for all xCnx \in \mathbb{C}^n.

Proof: By contradiction. Let AA be positive definite Hermitioan, Then λi>0 i=1,2,,σ\lambda_i > 0 \ \forall i = 1,2,\cdots,\sigma.

\rightarrow Suppose not, Let e0e \leq 0 be an eigenvector of AA.

From the positive definiteness of AA, we have e,Ae>0\langle e,Ae \rangle > 0.

e,Ae=e,λe=λe,e=λe,e>0\langle e,Ae \rangle = \langle e,\lambda e \rangle = \lambda^* \langle e,e \rangle =\lambda \langle e,e \rangle > 0

λe,e>0    λ>0\lambda \langle e,e \rangle > 0 \implies \lambda > 0 which is a contradiction. \blacksquare


#EE501 - Linear Systems Theory at METU